### Discussion :: Surveying - Section 1 (Q.No.7)

S.Venkata Siva said: (Apr 22, 2015) | |

Give me description. |

S.Venkata Siva said: (Apr 22, 2015) | |

Give me description. |

Jaswant Nayal said: (Jul 30, 2015) | |

Please guys give description. |

Pj_Baba said: (Oct 27, 2015) | |

Answer = (S/2r)*180*60/pi = 1718.87 S/R. |

Josh said: (Oct 29, 2015) | |

I don't think this answer (E) is correct. It is only correct if the sub chord, S, was actually the sub arc length. Otherwise, the real answer should be sin^-1 (S/2R)*60. |

Prakash Barik said: (Jan 27, 2016) | |

Please give some description my friend. |

Hazem said: (Mar 29, 2016) | |

Yes Josh, you are completely right. |

Umair said: (Aug 23, 2016) | |

Yes, E is 100% correct answer. |

Prem said: (Aug 24, 2016) | |

Please describe the question briefly. |

Abhishek Kumar said: (Sep 1, 2016) | |

Please describe it clearly. |

Mohammed Salahuddin Tirandaz said: (Nov 10, 2016) | |

Not understanding, Please describe it clearly. |

Yamanoor said: (Dec 23, 2016) | |

Angle of deflection=S/2R radians. = 180 * S/2 * R degree, = 180 * 60 * S/2 R* minutes, = 1718.9S/R. |

Naz said: (Jan 20, 2017) | |

Angle of reflection = 180 * s * 60/2R degree. = 1718.9S/R. |

Gman said: (Mar 7, 2017) | |

The angle of deflection = ((180/π) * s * 60)/(2 * R)°. |

Daxa Rathod said: (May 30, 2017) | |

Give me the description of this answer. |

Ved Prakash said: (Nov 25, 2017) | |

Actually the angular method of curve setting ,by Rankine's deflection method the deflection angle b/w chord and point of tangency is =1720C/R i, where c is the first sub chord and R is the radius. |

Amal Zaf Bannu said: (Dec 9, 2017) | |

E is the correct answer. I agree with the given answer. |

Shrikant said: (Jan 11, 2018) | |

Angle of deflection= (360/4π)* S/R. This gives answer 28.64 S/R. which is in degrees,, soo one degree is 60 min. Thus, 28.64 S/R degrees in radians is answer E) 1718.9 S/R. |

Williams said: (Apr 15, 2018) | |

Can anyone explain the solution in detail? |

Sameer Sopori said: (May 1, 2018) | |

@Williams. For an angle of deflection in RADIAN then use =S/2R. For an angle of deflection in DEGREE then use =(S/2R)* (180/π), For an angle of deflection in MINUTES then use = (S/2R)*(180/π)* 60. |

Nancy said: (Mar 6, 2019) | |

@Sameer. Thankyou for the explanation. |

Sai Adithya said: (Mar 7, 2019) | |

Thanks @Sameer Sopori. |

Govarthanan R said: (Jan 24, 2020) | |

For an angle of deflection in MINUTES then use = (S/2R)*(180/π)* 60. = S/R*((180*60)/(2*(22/7)). =S/R *1718.18, =1718.18S/R. |

Cemti said: (Mar 10, 2021) | |

@Govarthan. Please explain how did you get the 22/7? |

Nadeem said: (Mar 30, 2021) | |

Angel of deflection in radian = S/2R. And in radian =180 * S/2R * °. And in minute 180 * 60 * S/2R*°. After simplification =1718.87S/R. |

Karyeija Felex said: (Jul 27, 2021) | |

For all practical purposes sub chord is equal to sub arc length, so E is correct. |

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